Let a,n be positive integers, and let p be a prime. Prove if
(2^p) + (3^p) = (a^n), that n=1
2^5+3^5=275
275=5*5*11 , n>1
i proved for any prime p>5, n=1
Basically the proof runs like this: LHSmod(25)=0 iff the its last two digits {00,25,50,75}. Define B={00,25,50,75}. Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Consider f(k)=LT(2^k+3^k) for k=1,2,3,... Then we find f(k) B iff k=20m+5 or k=20m+15, for m=0,1,2,3,... Hence k cannot be a prime if LHSmod(25)=0. Hence k>5 cannot be a prime if n>2 on the RHS.
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EDIT: RED LETTERS
Now for the claim f(k) B iff k=20m+5 or k=20m+15, for m=0,1,2,3,... made above. See the attachment. All should be clear
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Edit: the smallest loop for LT is of length 20.
EDIT:
In the proof, (2^p) + (3^p) = (a^n) n=1 is true if {p}={odd numbers}-{20m+5}-{20m+15} because for these p LHSmod(25) 0. So the set of p's for which n=1 is true is strictly larger than the prime set.
Actually, I suspect that n=1 is true for ANY positive integer p! However a proof for that might be much more involved and I shall not attempt to prove it
Of course this proves the question!! and more than the question!
Let me try to clarify my reasoning a little for you:
1) for p=odd. We've shown this, right?
2) Now RHS If RHS=LHS, then RHS=0mod(5). If n>2, we must also have RHS=0mod(25). So LHS=0mod(25). So if we can show LHS 0mod(25) for prime p's, then we are done (this is also what abhishekkgp claimed up there).
3) But x=0mod(25) iff last two digits of x are 00, 25, 50 or 75. But LHS has those last two digits iff p=20m+5 or p=20m+15, which are NOT primes! Hence prime for prime p LHS and we're done!
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Why the table is enough? Because there's only FINITE p's we need to check (because we only care about last two digits!)
EDIT: Red letters
Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Then LT(2^5+3^5)=75. Define to be the smallest integer>5 such that 1) LT(2^ +3^ )=00 or 25 or 50 or 75; 2) LT(2^ )=LT(2^5); 3) LT(3^ )=LT(3^5). Such a is found by inspecting the first 20 terms of the table to be . And LT(2^ +3^ )=75. No problem with this?
Now because LT function takes only the last two digits, it is true that LT(2^(5+i))=LT =LT(LT(2^5)LT(2^i))=LT(LT(2^ )LT(2^i))=LT( )=LT( ), for any i>0. Similarly LT(3^(5+i))=LT( ), too. So LT(2^(5+i)+3^(5+i))=LT(LT(2^(5+i))+LT(3^(5+i)))=LT (LT(2^( ))+LT(3^( )))=LT(2^( +i)+3^( +i))
From this we can see f(k)=LT(2^k+3^k) takes only FINITELY MANY values for k=5,6,7,...
and the values it can take are exactly equal to this set:{f(5),f(6),f(7),...,f( )}, and any f(k) {00,25,50,75} iff there exists 5 k' such that f(k')=f(k).
So what is this k' if f(k') {00,25,50,75}? Inspect the table we find k'=5 or k'=15.
So now you see where k=5+20m and k=15+20m come from.
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If this doesn't convince you. I happily quit
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EDIT: red letters.